Abby is constructing the inscribed circle for △JKL .

Mathematics

2 answers:

Nady [450]2 years ago

5 0

Just took quiz. Correct answer is B

alex41 [277]2 years ago

3 0

Answer:

Hence, the correct answer is:

Construct the angle bisector of ∠J .

Step-by-step explanation:

The steps for the construction of inscribed circle are:

Bisect one of the angles

Bisect another angle

Where they cross is the center of the inscribed circle, called the incenter

Construct a perpendicular from the center point to one side of the triangle

Place compass on the center point, adjust its length to where the perpendicular crosses the triangle, and draw your inscribed circle!.

The answer that best suits the construction among the given option is:

Construct the angle bisector of ∠J.

(Since first steps of the construction will be satisfied)

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nataly862011 [7]

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(3x+2)(2x-5)(x³ -9)

Step-by-step explanation:

I like to use a graphing calculator to assist with factoring higher-degree polynomials. This polynomial has roots at -2/3, +5/2, and an irrational number near 2.08. The factors corresponding to the rational roots are (3x +2) and (2x -5).

Polynomial long division or synthetic division can be used to divide these out.

In each case, the divisor has a coefficient other than 1, so both dividend and divisor polynomials need to be divided by that coefficient before proceeding with synthetic division. (That is why there are fractions in the tableaux.)

The result is that the factorization in integers is ...

$6x^5-11x^4-10x^3-54x^2+99x+90=\bf{(3x+2)(2x-5)(x^3-9)}$

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The irrational roots are the three cube roots of 9, two of which are complex.

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Any of several different algorithms can tell you that the real roots will lie in the interval [-6, 6] or smaller. The rational roots will be of the form ...

±(divisor of 90)/(divisor of 6)